The smallest number of nodes in **a binary search tree** with a height of h is h+1 (in the case of left-skewed and right-skewed binary search trees). When all levels of a binary search tree are entirely full, the maximum number of nodes is h. There will be a total of 20 + 21 +... nodes. 2h is equal to 2(h+1)-1. Therefore, the number of nodes in a binary search tree with a height of h is 2(h+1)-1.

- What is the minimum and maximum number of nodes in a complete binary tree of height h?
- What is the height of the binary search tree?
- What is the maximum number of elements at height h in a binary tree whose root is at height 0?
- Which expression gives the maximum number of nodes?
- How many minimum nodes are needed to create a tree?
- What is the minimum and maximum number of nodes possible on a level 5 binary search tree?
- What is the maximum number of nodes?
- How many leaf nodes are present in a binary tree having a depth of H?

If a binary search tree has **n nodes**, the greatest height of the tree is n-1 and the least height is floor (log2n). There are also balanced binary search trees that contain 2^(h-1) + 1 nodes.

2 A binary tree of height 'h' can have a maximum of 2h-1 nodes. In this context, a tree's height is defined as the number of nodes on its root-to-leaf route. The height of a tree with a single node is taken to be 1. Thus, the maximum number of elements at height 'h' in a binary tree whose root is at height 0 is 2^h - 1.

At level 'l,' the maximum number of nodes is 2l-1. The number of nodes on the path from root to node, including the root itself, is denoted as level. We are assuming that the root level is 1. A binary tree of **height h** can have a maximum of **2h-1 nodes**.

In general, a binary tree with n levels has **2^n-1 nodes**.

Note that in practice, a binary search tree will not have exactly this many nodes; instead it will be a balanced tree, which we discuss in the next question.

Each of **the first two h levels** must have at least one node. At the first h layers, all potential nodes are present. A complete binary tree with h nodes contains 2h minus 1 nodes. 4 entire binary trees in height.

(b) with **5 nodes** has a height of floor (log25) = 2. The smallest number of nodes in a binary tree with a height of h is h+1 (in the case of left-skewed and right-skewed binary trees). Therefore, the smallest level-5 binary search tree contains 5 nodes and has a height of 4.

Every node in a binary tree has either no node, one node, or two nodes. As a result, we may argue that one node can only have a maximum of two nodes. The binary tree's leveling begins at level 0. At this level, the binary tree begins with a single node known as the root of the binary tree. This root node controls where the path from the trunk to the first leaf starts and ends. Because there is only one node at level 0, it cannot have any children.

Binary trees can contain an arbitrary number of nodes per level. Let's assume for a moment that the binary tree has three levels with **four, five, six, and seven nodes** respectively. Then the total number of nodes would be 24. This means that every node can have at most two other nodes. However, some nodes may only have one other node while others may have zero or more than two other nodes.

It is important to understand that the maximum number of nodes in **a binary tree** is equal to the depth of the tree. The depth of a tree is the shortest path from the root to a leaf. In **our example tree**, the depth is equal to its third level because that is the shortest distance between the root and a leaf (i.e., the last node on each branch). Trees with greater depths will have **more than 24 nodes**.

Trees with fewer than 24 nodes usually have exactly 24 nodes.

There are two leaf nodes in a perfect binary tree of height h. For the height, we'll utilize induction once more. There are 20 nodes when h = 0, and one of them is a leaf node. 2*2h = 2h + 1 leaf node is required for h + 1.